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In-the-sequence-of-numbers-1-2-11-22-111-222-the-sum-of-the-digits-in-999th-terms-is-

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Question Number 42494 by Tawa1 last updated on 26/Aug/18
In the sequence of numbers 1, 2, 11, 22, 111, 222, ... the sum of the digits in 999th terms is ??
$$\mathrm{In}\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{of}\:\mathrm{numbers}\:\:\mathrm{1},\:\mathrm{2},\:\mathrm{11},\:\mathrm{22},\:\mathrm{111},\:\mathrm{222},\:…\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits} \\ $$$$\mathrm{in}\:\mathrm{999th}\:\mathrm{terms}\:\mathrm{is}\:?? \\ $$
Commented by Tinku Tara last updated on 22/Apr/21
1st term 1 3rd term 11 5th term 111 7th term 1111 9th term 11111 (2n+1) term (n+1)time 1. 999=2×499+1 999 term will have 500 1s and sum of digits=500
$$\mathrm{1st}\:\mathrm{term}\:\mathrm{1} \\ $$$$\mathrm{3rd}\:\mathrm{term}\:\mathrm{11} \\ $$$$\mathrm{5th}\:\mathrm{term}\:\mathrm{111} \\ $$$$\mathrm{7th}\:\mathrm{term}\:\mathrm{1111} \\ $$$$\mathrm{9th}\:\mathrm{term}\:\mathrm{11111} \\ $$$$\left(\mathrm{2n}+\mathrm{1}\right)\:\mathrm{term}\:\left(\mathrm{n}+\mathrm{1}\right)\mathrm{time}\:\mathrm{1}. \\ $$$$\mathrm{999}=\mathrm{2}×\mathrm{499}+\mathrm{1} \\ $$$$\mathrm{999}\:\mathrm{term}\:\mathrm{will}\:\mathrm{have}\:\mathrm{500}\:\mathrm{1s} \\ $$$$\mathrm{and}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{digits}=\mathrm{500} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 28/Aug/18
the given series is combinationof two series serids are (1,11,111,...) and (2,22,222,....) first series (1,11,111,1111...) {(10^0 +1),(10^1 +1),(10^2 +10^2 +1),(10^3 +10^2 +10+1)...} 999th term 10^(999−1) +10^(997) +10^(996) +...+1=T_(999) T_(999) =((10^(998) (1−(1/(10^(999) ))))/(1−(1/(10))))=((10^(999) (1−(1/(10^(999) ))))/9) T_(999) =((10^(999) −1)/9)=f(10)
$${the}\:{given}\:{series}\:{is}\:{combinationof}\:{two}\:{series} \\ $$$${serids}\:{are}\:\left(\mathrm{1},\mathrm{11},\mathrm{111},…\right)\:{and} \\ $$$$\left(\mathrm{2},\mathrm{22},\mathrm{222},….\right) \\ $$$${first}\:{series}\:\left(\mathrm{1},\mathrm{11},\mathrm{111},\mathrm{1111}…\right) \\ $$$$\left\{\left(\mathrm{10}^{\mathrm{0}} +\mathrm{1}\right),\left(\mathrm{10}^{\mathrm{1}} +\mathrm{1}\right),\left(\mathrm{10}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} +\mathrm{1}\right),\left(\mathrm{10}^{\mathrm{3}} +\mathrm{10}^{\mathrm{2}} +\mathrm{10}+\mathrm{1}\right)…\right\} \\ $$$$\mathrm{999}{th}\:{term} \\ $$$$\mathrm{10}^{\mathrm{999}−\mathrm{1}} +\mathrm{10}^{\mathrm{997}} +\mathrm{10}^{\mathrm{996}} +…+\mathrm{1}={T}_{\mathrm{999}} \\ $$$${T}_{\mathrm{999}} =\frac{\mathrm{10}^{\mathrm{998}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}^{\mathrm{999}} }\right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}}}=\frac{\mathrm{10}^{\mathrm{999}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}^{\mathrm{999}} }\right)}{\mathrm{9}} \\ $$$$ \\ $$$${T}_{\mathrm{999}} =\frac{\mathrm{10}^{\mathrm{999}} −\mathrm{1}}{\mathrm{9}}={f}\left(\mathrm{10}\right) \\ $$$$ \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 27/Aug/18
check the answer
$${check}\:{the}\:{answer}\: \\ $$
Commented by Tawa1 last updated on 27/Aug/18
Oh, God bless you sir. I don′t know the answer. I will just write your solution sir. is the answer whole number or fraction if you solve further
$$\mathrm{Oh},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{the}\:\mathrm{answer}.\:\mathrm{I}\:\mathrm{will}\:\mathrm{just}\:\mathrm{write}\:\mathrm{your}\: \\ $$$$\mathrm{solution}\:\mathrm{sir}.\:\:\mathrm{is}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{whole}\:\mathrm{number}\:\mathrm{or}\:\mathrm{fraction}\:\mathrm{if}\:\mathrm{you}\:\mathrm{solve} \\ $$$$\mathrm{further} \\ $$

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